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Prove that

\begin{equation*} \frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\cdots +\frac{1}{x_n}=1,~∀i,x_i\in \mathbb{Z^+} \end{equation*}

has a finite number of integer solutions. I tried to solve $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ in integers and tried to deduce a solution for $\frac{1}{x}+\frac{1}{y}=1.$ And apply induction with the induction step that the number of solutions is finite from $n=1$ to $m-1$ and from there construct a solution for $n=m.$ I found

\begin{equation*} \frac{1}{(z-1)x}+\frac{1}{(z-1)y}= \frac{1}{z} \end{equation*}

but I rather had $1$ there in stead of $\frac{1}{z}.$ Stuck now. Help impossible for me?

Salomo
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user148584
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    I would first sort the solutions such that $x_1\le x_2\le\ldots \le x_n$. Clearly we must have $x_1\le n$, giving us an upper bound for $x_1$. Then for a fixed choice of $x_1$ we similarly get an upper bound for $x_2$ et cetera. – Jyrki Lahtonen May 05 '15 at 20:32
  • In other words, you might have better luck proving by induction on $n$ that the equation $$\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n}=q_n$$ has only finitely many integer solutions no matter which positive rational number $q_n$ is. – Jyrki Lahtonen May 05 '15 at 20:38
  • you mean for fixed $n$ right? – MichaelChirico May 05 '15 at 23:02
  • @MichaelChirico No, he meant $x_1$. Add @\user (without \ ) when you want to reply if you want him to receive a notification. – user26486 May 05 '15 at 23:21

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